The curves \(C_1\) and \(C_2\) are defined by \[y=e^{-x}\qquad (x>0)\qquad \text{and}\qquad y=e^{-x}\sin x \qquad (x>0),\] respectively. Sketch roughly \(C_1\) and \(C_2\) on the same diagram.
For \(C_1\) and \(C_2\), we need the parts of these curves that satisfy \(x > 0\).
The two curves \(\,y=e^{-x}\) and \(\,y=-e^{-x}\) form what is known as an `envelope’ around \(\,y=e^{-x}\sin x\).
Let \(x_n\) denote the \(x\)-coordinate of the \(n\)th point of contact between the two curves, where \(0<x_1< x_2<\dotsb\), and let \(A_n\) denote the area of the region enclosed by the two curves between \(x_n\) and \(x_{n+1}\). Show that \[A_n=\frac{1}{2}\left(e^{2\pi}-1\right)e^{-(4n+1)\pi/2}\]
First we need to know what the points \(x_n\) are. Solving \[e^{-x}=e^{-x}\sin x,\qquad \left(e^{-x}\neq 0 \right)\]\[\Longrightarrow \sin x = 1 \qquad \Longrightarrow \qquad x_1=\frac{\pi}{2},\, x_2=2\pi+\frac{\pi}{2} \dotsb\] Thus \[x_n=2\pi(n-1)+\frac{\pi}{2}=\frac{1}{2}(4n-3)\pi\] and \[A_n=\int_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi}\,e^{-x}\left(1-\sin x\right)\,dx,\] the difference between the two curves \(C_1\) and \(C_2\). This can be seen below.
Consider \(I=\int e^{-x}\sin x\,dx\). We can solve this via integration by parts, with \(u=\sin x\) and \(v'=e^{-x}\):
\[\begin{align*}
I &= \int e^{-x}\sin x \,dx \\
&= \left[-e^{-x}\sin x \right] + \int e^{-x}\cos x \,dx \\
&= \left[-e^{-x}\sin x \right] + \left[-e^{-x}\cos x \right] - \int e^{-x}\sin x \,dx \, \, \text{ (by integrating the integral above by parts again)}\\
&= \left[-e^{-x}\sin x \right] + \left[-e^{-x}\cos x \right] - I\\
\Longrightarrow \quad 2I &= -e^{-x}\left(\sin x+ \cos x\right)\\
\Longrightarrow \quad I &= -\frac{1}{2}e^{-x}\left(\sin x+ \cos x\right)
\end{align*}\]
Thus
\[\begin{align*}
A_n &=\int_{x_n}^{x_{n+1}}\,e^{-x}\left(1-\sin x\right)\,dx \\
&= \int_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi} \,e^{-x}\,dx - \int_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi} \,e^{-x}\sin x\,dx \\
&= \left[-e^{-x} + \frac{1}{2}e^{-x}\left(\sin x+ \cos x\right)\right]_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi} \, \, \text{(recognising the second integral above as $I$)}\\
&= \left[\frac{1}{2}e^{-x}\left(\sin x+ \cos x-2\right)\right]_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi} \\
&= \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}\left[\sin \left(2n\pi+\frac{\pi}{2}\right) + \cos \left(2n\pi+\frac{\pi}{2}\right) -2\right] \\
&\quad - \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}e^{2\pi}\left[\sin \left(2(n-1)\pi+\frac{\pi}{2}\right) + \cos \left(2(n-1)\pi+\frac{\pi}{2}\right) -2\right].
\end{align*}\]
With \(\,\sin\left(2n\pi + \frac{\pi}{2}\right)=\sin\frac{\pi}{2}=1\) and \(\,\cos\left(2n\pi + \frac{\pi}{2}\right)=\cos\frac{\pi}{2}=0\) for all integers \(n\), we find
\[\begin{align*}
A_n &= \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}\left(1 + 0 -2 \right) - \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}e^{2\pi}\left(1 + 0 -2\right) \\
&= \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}\left(-1 +e^{2\pi} \right) \\
&= \frac{1}{2}\left(e^{2\pi} -1\right)e^{-\frac{1}{2}(4n+1)\pi},
\end{align*}\]
as required.
[…] and hence find \(\displaystyle\sum\limits_{n=1}^{\infty} A_n\).
Let \(\,A=\displaystyle\sum\limits_{n=1}^{\infty} A_n\). Then
\[\begin{align*}
A &= \sum\limits_{n=1}^{\infty} \frac{1}{2}\left(e^{2\pi} -1\right)e^{-\frac{1}{2}(4n+1)\pi} \\
A &= \frac{1}{2}\left(e^{2\pi} -1\right) \sum\limits_{n=1}^{\infty} e^{-2n\pi}e^{-\frac{\pi}{2}} \\
A &= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{\pi}{2}} \sum\limits_{n=1}^{\infty} (e^{-2\pi})^n \\
A &= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{\pi}{2}} \left[e^{-2\pi} + (e^{-2\pi})^2 + (e^{-2\pi})^3 + \dotsb\right]\\
A &= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{\pi}{2}} e^{-2\pi} \left[1+ e^{-2\pi} + (e^{-2\pi})^2 + \dotsb\right]
\end{align*}\]
We therefore have the infinite sum of a geometric series, with first term \(\,a=\frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{5\pi}{2}}\) and ratio \(\,r=e^{-2\pi}\). Since \(\,2\pi > 0\), \(e^{-2\pi}<1\) and the series converges, with infinite sum \(\,S_{\infty}=\frac{a}{1-r}\). Therefore
\[\begin{align*}
A &= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{5\pi}{2}}\times \frac{1}{1-e^{-2\pi}} \\
&= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{5\pi}{2}} \times \frac{e^{2\pi}}{e^{2\pi}-1} \\
&= \frac{1}{2}e^{-\frac{5\pi}{2}} \times e^{2\pi} = \frac{1}{2}e^{-\frac{\pi}{2}}.
\end{align*}\]
So \(\displaystyle\sum\limits_{n=1}^{\infty} A_n = \frac{1}{2}e^{-\frac{\pi}{2}}\).
